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Series Contest (SC4)

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Post Thu May 15, 2014 6:10 am
Shobhit Site Admin
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Posts: 850
Location: Jaipur, India

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Welcome! I announce the Beginning of another Series Contest (SC4)! I am not going to post rules again and again as most of our members are familiar with the idea of contests. For newcomers I would suggest to go through the previous contests before taking part.

Remember numbering your problems while posting. Have fun and enjoy! :D

Post Thu May 15, 2014 6:12 am
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Problem 1

Show that \(\displaystyle \sum_{k=1}^\infty \text{Ci}(2\pi k n )=\frac{1}{2}\left\{\psi_0(n)-\log(n)+\frac{1}{2n} \right\}\)

where \(\displaystyle \text{Ci}(x)=-\int_{x}^{\infty}\frac{\cos(t)}{t}dt\)

Post Fri May 16, 2014 8:56 am
Random Variable Integration Guru
Integration Guru

Posts: 381
$ \displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = - \sum_{k=1}^{\infty} \int_{2 \pi k n}^{\infty} \frac{\cos t}{t} \ dt$

$\displaystyle = - \sum_{k=1}^{\infty} \int_{2 \pi kn}^{\infty} \int_{0}^{\infty} \cos t e^{-tx} \ dx \ dt = - \sum_{k=0}^{\infty} \int_{0}^{\infty} \int_{2 \pi k n}^{\infty} \cos t e^{-xt} dt \ dx$

$ \displaystyle = -\sum_{k=1}^{\infty} \int_{0}^{\infty} e^{-2 \pi k n x} \frac{x}{1+x^{2}} \ dx $

$ = \displaystyle -\int_{0}^{\infty} \frac{x}{1+x^{2}} \sum_{k=1}^{\infty} e^{-2 \pi k n x} \ dx = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{e^{- 2 \pi n}}{1-e^{-2 \pi n}} \ dx$

$ \displaystyle = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{e^{2 \pi nx}-1} \ dx $


Without the $n$, all we would need to do is differentiate Binet's log gamma formula.

If someone would like to continue from here, go ahead. Otherwise I'll try and finish it later.

Post Fri May 16, 2014 7:06 pm
Random Variable Integration Guru
Integration Guru

Posts: 381
I'm stupid. All you need to do is make a substitution.

$\displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = -\int_{0}^{\infty} \frac{x}{1+x^{2}} \frac{1}{e^{2 \pi nx}-1} \ dx$

$ \displaystyle = -\int_{0}^{\infty} \frac{u}{n^{2}+u^{2}} \frac{1}{e^{2 \pi u}-1} \ du$


And differentiating Binet's formula,

$ \displaystyle \psi(z) = \ln z + \left(z- \frac{1}{2} \right) \frac{1}{z} - 1 - 2 \int_{0}^{\infty} \frac{t}{z^{2}+t^{2}} \frac{1}{e^{2 \pi t}-1} \ dt$


Therefore,

$\displaystyle \sum_{k=1}^{\infty} \text{Ci}(2 \pi k n) = \frac{1}{2} \left( \psi(n) - \ln(n) + \frac{1}{2n} \right)$

Post Fri May 16, 2014 7:59 pm
galactus User avatar
Global Moderator
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Posts: 902
Very good G man :)

That's a nice way to go about it.

Here is a formula that may come in handy here or there:

it appears to be a Binet type formula

$$\psi(a)=log(a)-\frac{1}{2a}+2\sum_{n=1}^{\infty}\left[\cos(2\pi na)Ci(2\pi na)+\sin(2\pi na)si(2\pi na)\right]...(1)$$

Let $a=n$ and get:

$$\psi(n)=log(n)-\frac{1}{2n}+2\sum_{m=1}^{\infty}Ci(2n\pi m)$$



(1) comes about by using the Binet formula $$log\Gamma(u)=(u-1/2)log(u)-u+1/2log(2\pi)+\frac{1}{\pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\sin(2\pi nx)}{n(x+u)}dx$$

and diffing w.r.t u.

an interesting side note is:

$$2\int_{0}^{\infty}\frac{\tan^{-1}(x/u)}{e^{2\pi x}-1}dx=\frac{1}{\pi}\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{\sin(2\pi nx)}{n(x+u)}dx$$

The Abel-Plana sum formula can be a handy thing(I used it for that other problem S posted).

By using $f(n)=\frac{1}{n+u)^{s}}$, then:

$$\zeta(s,u)=\sum_{n=0}^{\infty}\frac{1}{(n+u)^{s}}=\frac{1}{2u^{s}}+\frac{u^{1-s}}{s-1}+i\int_{0}^{\infty}\frac{(u+ix)^{-s}-(u-ix)^{-s}}{e^{2\pi x}-1}dx$$

But, $$(u+ix)^{-s}-(u-ix)^{-s}=r^{-s}(e^{-is\theta}-e^{is\theta})=\frac{2}{i(u^{2}+x^{2})^{s/2}}\sin(s\cdot \tan^{-1}(x/u))$$

diffing and so forth brings one around to the Binet thing. I am sure you fellas already know this, but it is a cool thing to keep in the toolbelt.

Post Sat May 17, 2014 2:05 am
Random Variable Integration Guru
Integration Guru

Posts: 381
Problem 2


$ \displaystyle \sum_{k=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}}
\left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$

Post Sat May 17, 2014 9:12 am
zaidalyafey Global Moderator
Global Moderator

Posts: 357
Random Variable wrote:
Problem 2


$ \displaystyle \sum_{n=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}}
\left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$


so the index of summation must be $k=1$ , right ?
Wanna learn what we discuss , see Book

Post Sat May 17, 2014 11:20 am
Random Variable Integration Guru
Integration Guru

Posts: 381
zaidalyafey wrote:
Random Variable wrote:
Problem 2


$ \displaystyle \sum_{n=1}^{\infty} \frac{(2k-1)^{4n+1}}{1+e^{(2k-1) \pi}} = \frac{1}{2} \frac{1}{\pi^{4n+2}}
\left(1-\frac{2}{2^{4n+2}}\right) (4n+1)! \zeta(4n+2) = \frac{2^{4n+1}-1}{8n+4} B_{4n+2}$


so the index of summation must be $k=1$ , right ?


Yes.

Post Sat May 17, 2014 12:31 pm

Posts: 32
the sum can be gotten from this eq:
$$\sum\limits_{n{\rm{ = }}0}^\infty {\frac{{\left( {n + \frac{1}{2}} \right)}}{{{{\left( {n + \frac{1}{2}} \right)}^4} + {a^4}}}} \frac{1}{{{e^{2\pi n + 1}} + 1}} = - \frac{\pi }{{2{a^2}}} \times \frac{1}{{\exp \left( {2\sqrt 2 a\pi } \right) + 2\exp \left( {\sqrt 2 a\pi } \right)\cos \left( {\sqrt 2 a\pi } \right) + 1}} + \frac{\pi }{{8{a^2}}} \\ \, \, \, - \frac{1}{{2{a^2}}}{\mathop{\rm Im}\nolimits} \psi \left( {\frac{1}{2} + \frac{a}{{\sqrt 2 }} + \frac{a}{{\sqrt 2 }}i} \right)$$
Generalized Abel-Plana Formula will work on it (reference :The generalized Abel-Plana formula with applications to Bessel functions and Casimir effect,Ramanujan'lost book),but Yelusalen gave another proof by using Complex Analysis,he considered the contour integral
$$\int_C {\frac{z}{{{z^4} + {a^4}}}\frac{1}{{{e^{2\pi z}} + 1}}} \frac{1}{{{e^{ - 2\pi iz}} + 1}}dz$$,the contour of which is $$[0,R] + {C_R} + [Ri,0],R \to \infty$$,bypassing all poles on the borders.
Last edited by zaidalyafey on Sat May 17, 2014 1:29 pm, edited 1 time in total.
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Post Mon May 19, 2014 11:22 am
Random Variable Integration Guru
Integration Guru

Posts: 381
Marko Riedel evaluated the second series using the inverse Mellin transform.

http://math.stackexchange.com/questions ... 188#740188

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